Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance

Question

Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the charges and moves along the line connecting them. What is the electric potential energy of the electron when it is

1. at the midpoint?
2. 10.0 cm from the 3.00 nC charge?

I tried to do the first part of this problem by observing another example, but I still got it all wrong. Could you please show me the step-by-steps of how to do this problem and how you came up with all the equations??? Thanks!

Explanation / Answer

Point charge =q1 = + 3.00*10^-9 C q2 = + 2.00*10^-9 C Separation by distance = s = 50.0cm =0.5 m. Electric potential energy of electron at 0.25 m =potential energy due to q1 + potential energy due to q2 Electric potential energy of electron at 0.25 m =-9*10^9(1.6*10^-19)[ 3.00*10^-9 + 2.00*10^-9 ] /0.25 Electric potential energy of electron at 0.25 m =-9*(1.6*10^-19) [ 3.00+ 2.00 ] /0.25 Electric potential energy of electron at 0.25 m = - 9*(1.6*10^-19)20 Electric potential energy of electron at 0.25 m = - 2.88*10^-17J Electric potential energy of electron at 0.10 m =-9*10^9(1.6*10^-19)*10^-9[ 3.00/0.1 + 2.00/0.4 ] Electric potential energy of electron at 0.10 m =-5.04 *10^ -17 J Change in PE =2.16*10^-17 J KE =2.16*10^-17 J Speed = v = sq rt [2KE/m ] Speed = v = sq rt [2*2.16*10^-17 /9.1*10^ -31 ] Speed = v = 6.890*10^6 m/s When electron is 10.0cm from the + 3.00-nC charge its speed is 6.890*10^6 m/s

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